An electrical appliance having a resistance of 200 Ω is operated at 200 V. Calculate the energy consumed by the appliance in 5 minutes (i) in joule, (ii) in kWh

(i) Given,

Resistance (R) = 200 Ω

Voltage (V) = 200 volt

Time (t) = 5 min = 300 sec

As we know,

Energy (E) = $\dfrac{V^2t}{R}$

Substituting the values in the formula above we get,

$E = \dfrac{200^2 \times 300}{200} \\[0.5em] \Rightarrow E = 60,000 \text{ J}$

Hence, energy consumed = 60,000 J

(ii) In kWh = ?

As we know,

$1 \text{kWh} = 3.6 \times 10^6 \text{J} \\[0.5em] 1 \text{ J} = \dfrac{1}{3.6 \times 10^6} \text{kWh} \\[0.5em] 60,000 \text{ J} = \dfrac{60,000}{3.6 \times 10^6} \\[0.5em] \Rightarrow 60,000 \text{ J}= 0.0167 \text { kWh} \\[0.5em]$

Hence, energy consumed = 0.0167 kWh